Download Poisson Distribution Examples

Lecture 15
Agenda
1. Poisson Distribution Examples
2. Hypergeometric Distribution
Poisson Distribution Examples
Example 1
The manager of a industrial plant is planning to buy a machine of
either type A or type B. For each day’s operation the number of repairs
X, that the machine A needs is a poisson random variable with mean
0.96. The daily cost of operating A is CA = 160 + 40 ∗ X 2 . For
machine B, let Y be the random variable indicating the number of
daily repairs, which has mean 1.12, and the daily cost of operating
B is CB = 128 + 40 ∗ Y 2 . Assume that the repairs take negligible
time and each night the machine are cleaned so that they operate
like new machine at the start of each day. Which machine minimizes
the expected daily cost ?
Solution. The expected cost for machine A is,
E(CA ) =
=
=
=
=
E(160 + 40 ∗ X 2 )
160 + 40 ∗ E(X 2 )
160 + 40 ∗ (V (X) + [E(X)]2 )
160 + 40 ∗ (0.96 + 0.962 )
235.264
The expected cost for machine B is,
E(CB ) =
=
=
=
=
E(128 + 40 ∗ Y 2 )
128 + 40 ∗ E(Y 2 )
128 + 40 ∗ (V (Y ) + [E(Y )]2 )
128 + 40 ∗ (1.12 + 1.122 )
222.976
1
Example 2
The number of calls coming per minute into a hotels reservation
center is Poisson random variable with mean 3.
(a) Find the probability that no calls come in a given 1 minute period.
(b) Assume that the number of calls arriving in two different minutes
are independent. Find the probability that atleast two calls
will arrive in a given two minute period.
Solution. (a) Let X denote the number of calls coming in that given 1
minute period.
X ∼ P oisson(3)
P (X = 0) =
e−3 30
= e−3
0!
(b) Let X1 and X2 be the number of calls coming in the first and second
minutes rspectively.
We want P (X1 + X2 ≥ 2).
P (X1 + X2 ≥ 2) = 1 − P (X1 + X2 < 2)
= 1 − [P (X1 = 0, X2 = 0) + P (X1 = 1, X2 = 0) + P (X1 = 0, X2 = 1)]
= 1 − [P (X1 = 0) ∗ P (X2 = 0) + P (X1 = 1) ∗ P (X2 = 0) +
P (X1 = 0) ∗ P (X2 = 1)]
31
31
= 1 − e−3 e−3 − e−3 e−3 − e−3 e−3
1!
1!
= 1 − e−6 − 3e−6 − 3e−6
= 1 − 7e−6
Hypergeometric Distribution
Example 3
Suppose 30% of the UF students are graduates. If I take a random
sample of 10 students find the probability that the number of graduates
is atmost 4.
Let X be the number of graduates among the 10 students. Then
X ∼ Bin(10, 0.3)
2
Using our usual formulas,
P (X ≤ 4) = 0.8497
Example 4
Suppose I have 50 UF students in my class and 30% of them are graduates.
If I take a random sample of 10 students from them find the probability
that the number of graduates is atmost 4.
Let X be the number of graduate students among the 10 students. In this
case we cannot say that X ∼ Bin(10, 0.3), because for that to happen we
need 10 independent Bernoulli trials. But here we have 50 students among
whom 15 are graduates and 35 are undergraduates. So, for the first trial
P (Graduate) =
15
= 0.3
15 + 35
and that went as expected.
But now suppose in the first trial I got a graduate student. So at the beginning of the second trial we have 14 graduates and 35 undergrads. So
now,
14
14
=
6= 0.3
P (Graduate) =
14 + 35
49
That should not happened if they were independent Bernoulli trials with
probability of success 0.3.
Thus we need Hypergeometric Distribution.
Definition 1. A population of size N contains two types of items, k items
of Type I and N − k items of Type II. We randomly sample n items from
N , without repitition. Let X denote the number of items of Type I in our
sample.
Then X follows Hypergeometric Distribution with parameters (N, k, n)
Range(X) = {x| max{0, n − N + k} ≤ x ≤ min{n, k}}
k N −k
P (X = x) =
3
x
n−x
N
n
For Example 4 we have, N = 50, k = 15 and n = 10. So now
Range(X) = {x| max{0, 10−50+15} ≤ x ≤ min{10, 15}} = {x|0 ≤ x ≤ 15}
35 4
15
X
x
10−x
P (X ≤ 4) =
= 0.8750
50
10
x=0
Since we have established the pmf we will head for mean and variance.
They can be calculated the hard way i.e. doing it algebrically. But we won’t
do that. Let’s learn a few things before doing those calculations.
Independence of discrete random variables
For any two events, A and B, we say A and B are independent, if
P (A ∩ B) = P (A) × P (B)
For the past few lectures we have been using the concept of independence of
random variables without really defining it.
Definition 2. Let X1 , X2 , . . . Xk be k discrete random variables. We say
that {X1 , X2 , . . . Xk } are independent if for any
x1 ∈ Range(X1 ), x2 ∈ Range(X2 ), . . . , xk ∈ Range(Xk )
P (X1 = x1 , X2 = x2 , . . . Xk = xk ) = P (X1 = x1 )×P (X2 = x2 )×. . .×P (Xk = xk )
We note that for i = 1, 2, . . . , k; {Xi = xi } is an event. The above definition says that those k events are independent.
Please note one other thing; when we defined a discrete random variable
X, we said it has a range like, Range(X) = {x1 , x2 , x3 , . . .}. But here
x1 , x2 , x3 , . . . don’t belong to range of one variable X; but x1 denotes some element of Range(X1 ), x2 denotes some element of Range(X2 ),and so on.Please
don’t get confused over this.
Next we state another theorem, we will prove it next class.
Theorem 1. Let {X1 , X2 , . . . Xk } be k independent random variables. Let
f1 (), f2 (), . . . , fk () be k functions.Then
E(f1 (X1 )×f2 (X2 )×. . .×fk (Xk )) = E(f1 (X1 ))×E(f2 (X2 ))×. . .×E(fk (Xk ))
4